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PROFESSOR NELSON: Over the last
couple lectures through our
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consideration of a number of
somewhat difficult topics,
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including these Carnot cycles
and specific Carnot engine to
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represent them, we've tried to
define and understand a little
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bit about this very special
function, this entropy.
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Where we've got our dS.
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It's dq reversible over T.
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And our treatment of reversible
and irreversible cyclic
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processes, what we saw is that
if we go around a cycle and
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look at dq reversible over
T, then this is zero.
19
00:01:18 --> 00:01:20
Now go all the way around a
cycle, of course, this is
20
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just delta S for the cycle.
21
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And that's consistent with
the idea that entropy
22
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is a state functions.
23
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So if we go around cycle the
cycle, we have the same ending
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point, the starting point, same
state, then the change in the
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state function has to be zero.
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00:01:39 --> 00:01:48
But we saw that for an
irreversible path around a
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cycle, then we have that, this
gives something less than zero.
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00:01:57 --> 00:02:01
And so that was expressed in
Clausius inequality that
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includes both these cases.
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And now I just want to
go through a number of
31
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calculations of entropy.
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How entropy changes for a
number processes, both to just
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learn more about it in general
terms and also just to see how
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it's calculated and what sort
of changes in undergoes for
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00:02:23 --> 00:02:25
different sorts of
changes in state.
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00:02:25 --> 00:02:31
So the first very general
example that I want to
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show is the following.
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Let's say we've got an isolated
system, and it undergoes
39
00:02:46 --> 00:02:57
some irreversible change.
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00:02:57 --> 00:03:08
So we'll start at one, and
we'll go irreversibly
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to some new state, two.
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00:03:19 --> 00:03:28
OK, now we can always return
the system back to the
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initial state reversibly.
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00:03:36 --> 00:03:40
If we do that, then we can't
keep the system isolated.
45
00:03:40 --> 00:03:45
So it won't be
isolated anymore.
46
00:03:45 --> 00:03:48
In other words, in order to
arrange things so that you
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could have a return along a
reversible path, it
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can't be isolated.
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00:03:53 --> 00:03:56
In some sense you know that,
because when it was isolated
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00:03:56 --> 00:03:59
it spontaneously irreversibly
went from one to two.
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00:03:59 --> 00:04:03
And example would be ice
melting at some temperature
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that might be higher
than the freezing point.
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So you've got solid ice, but
the temperature is above
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zero degrees Celsius, so
irreversibly it'll melt.
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Systems isolated, it'll melt.
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Then you could put the system
in contact with a cold
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00:04:17 --> 00:04:23
reservoir reversibly re-freeze
the water to form ice again.
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But of course to do that you
couldn't keep it isolated.
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You need to have it in contact
with a low temperature bath.
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But again, this is just
completely general
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at this point.
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It's some irreversible
process returned then
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along a reversible path.
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So let's just look
at what happens.
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So here's our path B.
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Along path A, well it's
an isolated system.
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So the heat exchanged is zero.
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And then, for the cycle based
on the Clausius inequality
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expressed here, the integral
around the cycle of dq over
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T in general, it's less
than or equal to zero.
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If there was an irreversible
process involved, then in fact,
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it'll be less than zero.
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I'll write it here in the more
general case, but we know what
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it's really going to be with
the irreversible step.
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Well, so we can write this as
the integral of dq irreversible
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over T, going from
state one to state two.
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Oh, let me rewrite
this down here.
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So around the cycle, dq over T,
it's integral from state one to
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two of dq irreversible over T,
plus the integral from two back
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00:06:20 --> 00:06:27
to one of dq reversible over T.
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And of course, we know
that that's less than
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or equal to zero.
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But this, we know is zero,
right, because this
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irreversible step is
taken in isolation.
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The system is isolated.
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00:06:40 --> 00:06:42
We've already seen that
there's no heat crossing
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the boundary of the system.
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So we just have this.
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00:06:48 --> 00:06:58
So of course this dq reversible
over T from two to one, that's,
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what now we've a reversible
path, so that's just the
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entropy at the delta S
going from two to one.
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00:07:04 --> 00:07:11
It's S1 minus S2 it's delta S,
we can write it as delta S
93
00:07:11 --> 00:07:15
backwards, right, we're
going back along that
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reversible path.
95
00:07:17 --> 00:07:23
Which is of course
negative delta S forward.
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00:07:23 --> 00:07:29
Because around the whole cycle
delta s has to be zero.
97
00:07:29 --> 00:07:43
So what that says is delta
S forward is greater
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00:07:43 --> 00:07:50
than or equal to zero.
99
00:07:50 --> 00:07:52
That's interesting.
100
00:07:52 --> 00:07:54
What that's saying is
-- remember, we didn't
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00:07:54 --> 00:07:55
specify the process.
102
00:07:55 --> 00:08:00
It's just for any completely
general irreversible process.
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00:08:00 --> 00:08:06
For any such process, for any
spontaneous process, this
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00:08:06 --> 00:08:08
is going to be the case.
105
00:08:08 --> 00:08:13
So in other words, for an
isolated system, delta S
106
00:08:13 --> 00:08:18
tells us the direction
of spontaneous change.
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00:08:18 --> 00:08:29
In particular, for the isolated
system, delta S is greater
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00:08:29 --> 00:08:40
than zero for something
that's irreversible.
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00:08:40 --> 00:08:51
Delta S equals zero for
something that's reversible,
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00:08:51 --> 00:09:02
and delta S is never
less than zero.
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00:09:02 --> 00:09:04
In a sense, this is a
direct consequence of
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00:09:04 --> 00:09:11
the Clausius inequality.
113
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That's a pretty useful result.
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00:09:17 --> 00:09:21
So remember when we started
this whole discussion, I tried
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00:09:21 --> 00:09:25
to emphasize that the first
law of thermodynamics told us
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00:09:25 --> 00:09:27
about conservation of energy.
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00:09:27 --> 00:09:30
But it didn't tell us which
way something would go
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00:09:30 --> 00:09:34
spontaneously if you just left
the system to its own devices
119
00:09:34 --> 00:09:36
under certain conditions.
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00:09:36 --> 00:09:40
This is telling us that.
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00:09:40 --> 00:09:43
We can tell which way the
system will evolve in
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00:09:43 --> 00:09:46
which direction it'll go.
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00:09:46 --> 00:09:49
This is for an isolated system
and we'll generalize this
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00:09:49 --> 00:10:00
in subsequent lectures.
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00:10:00 --> 00:10:08
Now, let's look at what happens
if the system isn't isolated.
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00:10:08 --> 00:10:25
So now let's go from one
to two not isolated.
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00:10:25 --> 00:10:32
So of course, delta S I'm going
to specify for the system
128
00:10:32 --> 00:10:37
for reasons that'll soon be
obvious is S2 for the system
129
00:10:37 --> 00:10:39
minus S1 for the system.
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00:10:39 --> 00:10:42
Of course it doesn't depend
on the path, even though to
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00:10:42 --> 00:10:52
calculate it, we need to
find a reversible path.
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00:10:52 --> 00:10:55
Just to add emphasis to what
we just did here, of course,
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00:10:55 --> 00:10:57
it was the same thing.
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00:10:57 --> 00:11:02
In order to calculate delta S
along that irreversible path
135
00:11:02 --> 00:11:05
for that irreversible process
we needed to construct
136
00:11:05 --> 00:11:09
some reversible path.
137
00:11:09 --> 00:11:13
And what we did is imagined the
reversible path going backwards
138
00:11:13 --> 00:11:16
and then said, OK, fine then
delta S for the irreversible
139
00:11:16 --> 00:11:18
process must be the opposite of
delta S that we calculated
140
00:11:18 --> 00:11:21
along the reversible
bath going backwards.
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00:11:21 --> 00:11:24
The point is that to calculate
delta S in either direction,
142
00:11:24 --> 00:11:27
we needed to construct
some reversible path.
143
00:11:27 --> 00:11:31
All right, and of course
it'll be the same here.
144
00:11:31 --> 00:11:35
Now, in this case since the
system isn't isolated, that
145
00:11:35 --> 00:11:38
means heat's going to flow
across the boundary.
146
00:11:38 --> 00:11:43
So that means heat is going
between, it's being exchanged
147
00:11:43 --> 00:11:45
between the system
and the surroundings.
148
00:11:45 --> 00:11:47
So what's happening
to the surroundings?
149
00:11:47 --> 00:11:49
Generally, we've worry a lot
about the system, but not
150
00:11:49 --> 00:11:54
much in recent discussions
about the surroundings.
151
00:11:54 --> 00:12:01
But delta S surroundings
must be something.
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00:12:01 --> 00:12:04
We could calculate it.
153
00:12:04 --> 00:12:07
And that's depends on the path.
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00:12:07 --> 00:12:10
Not because it isn't the state
function in the surroundings,
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00:12:10 --> 00:12:15
of course it is, but because
depending on the path, the
156
00:12:15 --> 00:12:22
final state of the surroundings
will be different.
157
00:12:22 --> 00:12:25
After all, depending on how
much heat got exchanged, the
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00:12:25 --> 00:12:30
surroundings got more
or less of that heat.
159
00:12:30 --> 00:12:33
Work might have been done
on the surroundings.
160
00:12:33 --> 00:12:37
So the surroundings are going
to respond, are going to
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00:12:37 --> 00:12:39
change also, when the
system changes in this way
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00:12:39 --> 00:12:41
and it's not isolated.
163
00:12:41 --> 00:12:44
And how the surroundings
change will depend on
164
00:12:44 --> 00:12:45
what path was taken.
165
00:12:45 --> 00:12:47
And you've seen that before
in calculating things like
166
00:12:47 --> 00:12:51
pressure, volume, work
along different paths.
167
00:12:51 --> 00:12:54
Where you calculated delta u
for the system and saw that it
168
00:12:54 --> 00:12:59
was the same for different
paths but q and w were not the
169
00:12:59 --> 00:13:01
same for the system, and
therefore, of course, not for
170
00:13:01 --> 00:13:07
the surroundings either.
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00:13:07 --> 00:13:21
So let's just consider
an irreversible path.
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00:13:21 --> 00:13:23
And in order to think about
what happens to the
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00:13:23 --> 00:13:29
surroundings, let's take the
entire universe which consists
174
00:13:29 --> 00:13:36
of the system and the
surroundings -- that is a big
175
00:13:36 --> 00:13:40
isolated system, because
it's everything.
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00:13:40 --> 00:13:45
So it's not somehow in contact
with another additional
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00:13:45 --> 00:13:48
body or universe.
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00:13:48 --> 00:13:54
We can treat the entire
universe as an isolated system.
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00:13:54 --> 00:14:05
So, we're going to think
about the entire universe.
180
00:14:05 --> 00:14:13
We can think about it
as an isolated system.
181
00:14:13 --> 00:14:16
Great.
182
00:14:16 --> 00:14:23
And we know that delta S for
an isolated system, for
183
00:14:23 --> 00:14:28
a spontaneous process,
is greater than zero.
184
00:14:28 --> 00:14:30
That's right there.
185
00:14:30 --> 00:14:38
So delta S for the universe
which is delta S for the
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00:14:38 --> 00:14:44
system plus delta S for the
surroundings, and I'll specify
187
00:14:44 --> 00:14:49
since we're considering an
irreversible process, specify
188
00:14:49 --> 00:14:52
that, it's greater than zero.
189
00:14:52 --> 00:14:55
The whole thing happened
spontaneously, and the whole
190
00:14:55 --> 00:14:58
universe is an isolated system.
191
00:14:58 --> 00:15:05
So if we wanted to, we can
write that delta S irreversible
192
00:15:05 --> 00:15:11
for the surroundings must be
bigger than minus delta S for
193
00:15:11 --> 00:15:19
the system in order that
there sum be positive.
194
00:15:19 --> 00:15:21
All right?
195
00:15:21 --> 00:15:33
Now, if we consider a
reversible process, then of
196
00:15:33 --> 00:15:37
course we already know what
happens for an isolated system
197
00:15:37 --> 00:15:43
in a reversible process,
delta S is zero.
198
00:15:43 --> 00:15:47
So delta S in this case of the
whole universe, which is delta
199
00:15:47 --> 00:15:53
S for the system plus delta S
for the surroundings in the
200
00:15:53 --> 00:16:02
reversible case must
be equal to zero.
201
00:16:02 --> 00:16:08
In other words, delta S of the
surroundings in the reversible
202
00:16:08 --> 00:16:15
case is exactly the opposite
of delta S of the system.
203
00:16:15 --> 00:16:18
Delta S of the system is the
same in either case because
204
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reversible or irreversible,
we're specifying the system,
205
00:16:21 --> 00:16:24
goes between the same states,
but what happens to the
206
00:16:24 --> 00:16:30
surroundings is different
in the two cases.
207
00:16:30 --> 00:16:36
So one way of recasting this
statement or the Clausius
208
00:16:36 --> 00:16:41
inequality is the entropy of
the whole universe
209
00:16:41 --> 00:16:44
never decreases.
210
00:16:44 --> 00:16:47
Of course, in some sense,
that's obvious from this
211
00:16:47 --> 00:16:49
statement, since we're
considering the universe
212
00:16:49 --> 00:16:51
an isolated system.
213
00:16:51 --> 00:16:56
So the entropy in any process
can, it can, the change in
214
00:16:56 --> 00:17:04
entropy can either be zero or
positive and never negative.
215
00:17:04 --> 00:17:10
216
00:17:10 --> 00:17:27
So entropy off the whole
universe never decreases.
217
00:17:27 --> 00:17:30
That's actually a really
quite profound conclusion.
218
00:17:30 --> 00:17:36
All sorts of philosophical and
other consequences of it that
219
00:17:36 --> 00:17:39
are sometimes considered in
things like evolutionary
220
00:17:39 --> 00:17:41
biology and other things.
221
00:17:41 --> 00:17:43
And how things can change
altogether an entire
222
00:17:43 --> 00:17:48
system of things.
223
00:17:48 --> 00:17:55
Now, I'd like to just go
through a few calculations of
224
00:17:55 --> 00:17:59
delta S for common processes.
225
00:17:59 --> 00:18:01
We've seen a few general
features of entropy.
226
00:18:01 --> 00:18:05
Let's calculate it
for a few things.
227
00:18:05 --> 00:18:15
So calculations of delta S.
228
00:18:15 --> 00:18:17
And we know the
general procedure.
229
00:18:17 --> 00:18:20
Whatever the process is,
we're going to need to find
230
00:18:20 --> 00:18:24
reversible paths, and along a
reversible path then, we'll
231
00:18:24 --> 00:18:28
be able to calculate it.
232
00:18:28 --> 00:18:39
One, let's just take two
pieces of material at
233
00:18:39 --> 00:18:43
different temperatures.
234
00:18:43 --> 00:18:56
The entire system is
going to be isolated.
235
00:18:56 --> 00:19:01
And then let's connect them.
236
00:19:01 --> 00:19:02
And then let's connect them.
237
00:19:02 --> 00:19:03
Put some material between them.
238
00:19:03 --> 00:19:05
So now heat can flow
from one to the other.
239
00:19:05 --> 00:19:07
And you know what's going to
happen which is if they start
240
00:19:07 --> 00:19:09
out at unequal temperatures,
and you wait a little while,
241
00:19:09 --> 00:19:10
eventually they're going
to wind up at the
242
00:19:10 --> 00:19:12
same temperature.
243
00:19:12 --> 00:19:14
In other words, you
know the direction of
244
00:19:14 --> 00:19:17
spontaneous change.
245
00:19:17 --> 00:19:23
So let's just see what
happens to delta S.
246
00:19:23 --> 00:19:33
So isolated system, so
there's no work being done.
247
00:19:33 --> 00:19:35
There's no heat that's
being exchanged.
248
00:19:35 --> 00:19:42
So delta u is zero.
249
00:19:42 --> 00:19:44
What's dS?
250
00:19:44 --> 00:19:49
Well it's just the sum of the
changes in entropy for the two
251
00:19:49 --> 00:19:53
sub-systems the two
separate pieces.
252
00:19:53 --> 00:20:03
So it's dq1 over T1 plus dq2
over T2, I think there's a
253
00:20:03 --> 00:20:11
type in your notes. it's
plus not minus of course.
254
00:20:11 --> 00:20:16
But in this case, dq1 is
just the opposite of dq2.
255
00:20:16 --> 00:20:20
Because whatever heat is
flowing into T1, into this
256
00:20:20 --> 00:20:23
block, is coming out of this
block, or vice versa, depending
257
00:20:23 --> 00:20:32
on which one is hotter.
258
00:20:32 --> 00:20:39
So this is just equal to dq1
times one over T1 minus one
259
00:20:39 --> 00:20:45
over T2, which is T2 minus
T1 over T1 times T2.
260
00:20:45 --> 00:20:49
261
00:20:49 --> 00:20:54
So that's dS, and then we could
integrate it to get the delta
262
00:20:54 --> 00:20:57
S, but this is sufficient for
what I'd like to illustrate,
263
00:20:57 --> 00:20:59
which is just the
sign of things.
264
00:20:59 --> 00:21:09
We know dS must be greater
than zero for the spontaneous
265
00:21:09 --> 00:21:14
change that's going to occur.
266
00:21:14 --> 00:21:18
So let's just make sure that
common sense prevails here.
267
00:21:18 --> 00:21:23
That says if T2 is greater
than T1, right T2 is hotter.
268
00:21:23 --> 00:21:27
Then dq must be positive, so
this is positive, this is
269
00:21:27 --> 00:21:30
positive, dS we know
has to be positive.
270
00:21:30 --> 00:21:35
So, it says T2 is hotter.
271
00:21:35 --> 00:21:40
This being positive means heat
is flowing in from this one to
272
00:21:40 --> 00:21:45
this one, from the hotter
to the colder body.
273
00:21:45 --> 00:21:47
So it looks right.
274
00:21:47 --> 00:21:52
If T2 is lower than T1, if this
one is colder, then in order
275
00:21:52 --> 00:21:55
for dS to be positive,
which it has to be, this
276
00:21:55 --> 00:21:57
better be negative.
277
00:21:57 --> 00:22:00
In other words, again heat
is going to flow from the
278
00:22:00 --> 00:22:05
hotter toward the colder,
into the colder body.
279
00:22:05 --> 00:22:11
So, as promised, the condition
that dS must be greater than
280
00:22:11 --> 00:22:17
zero, guides us and tells us
the direction that spontaneous
281
00:22:17 --> 00:22:21
change is going to occur.
282
00:22:21 --> 00:22:23
We can tell just from
this condition which way
283
00:22:23 --> 00:22:44
things have to evolve.
284
00:22:44 --> 00:22:48
All right, let's try another
pretty simple process.
285
00:22:48 --> 00:22:59
Let's just take a gas in some
volume V and over here is
286
00:22:59 --> 00:23:05
going to be vacuum of equal
volume, and we're going
287
00:23:05 --> 00:23:07
to remove the barrier.
288
00:23:07 --> 00:23:10
You know what's going to
happen spontaneously, right?
289
00:23:10 --> 00:23:14
The gas is going to fill
the available volume once
290
00:23:14 --> 00:23:16
it becomes available.
291
00:23:16 --> 00:23:23
So this, we'll make this a
Joule expansion, and we'll
292
00:23:23 --> 00:23:29
make it an ideal gas.
293
00:23:29 --> 00:23:37
So the process is one mole of
gas at our initial volume and
294
00:23:37 --> 00:23:46
some temperature, and we'll
make this adiabatic, and this
295
00:23:46 --> 00:23:53
goes to one mole of gas at a
new volume, 2V and at
296
00:23:53 --> 00:23:55
the same temperature.
297
00:23:55 --> 00:24:01
All right, so there's
no work done.
298
00:24:01 --> 00:24:02
It's adiabatic.
299
00:24:02 --> 00:24:03
There's no heat exchanged.
300
00:24:03 --> 00:24:08
Delta u is zero.
301
00:24:08 --> 00:24:10
Everything's zero.
302
00:24:10 --> 00:24:13
Almost everything is zero.
303
00:24:13 --> 00:24:16
What isn't zero?
304
00:24:16 --> 00:24:16
STUDENT: Entropy.
305
00:24:16 --> 00:24:17
PROFESSOR NELSON: Entropy.
306
00:24:17 --> 00:24:19
It's going to tell us which
way this whole thing goes.
307
00:24:19 --> 00:24:22
So this is an
irreversible process.
308
00:24:22 --> 00:24:26
In order to calculate delta
S we need to construct a
309
00:24:26 --> 00:24:30
reversible path along
which this can go.
310
00:24:30 --> 00:24:34
So here's a way to do it.
311
00:24:34 --> 00:24:38
We could compress it back,
isothermally and reversibly.
312
00:24:38 --> 00:24:41
Just like the example I gave,
the general example, where I
313
00:24:41 --> 00:24:42
mentioned the ice melting.
314
00:24:42 --> 00:24:44
In that case, it
won't be adiabatic.
315
00:24:44 --> 00:24:48
We'd have to put it in
contact with a heat
316
00:24:48 --> 00:24:51
source of some sort.
317
00:24:51 --> 00:25:00
So reversible process.
318
00:25:00 --> 00:25:21
We could compress it back
to volume V, isothermally,
319
00:25:21 --> 00:25:22
and reversibly.
320
00:25:22 --> 00:25:25
Let me say, there's no -- we
wouldn't have to consider
321
00:25:25 --> 00:25:26
this process in reverse.
322
00:25:26 --> 00:25:29
Of course, we could also
consider the forward process
323
00:25:29 --> 00:25:34
in the presence of some heat
source with isn't isolated and
324
00:25:34 --> 00:25:37
do this, but let's just
consider the compression.
325
00:25:37 --> 00:25:46
So, OK, now we'll have
the -- well, once
326
00:25:46 --> 00:25:48
again, delta u is zero.
327
00:25:48 --> 00:25:50
It's an ideal gas.
328
00:25:50 --> 00:25:52
There's no temperature change.
329
00:25:52 --> 00:25:55
But this time q and w
aren't going to be zero.
330
00:25:55 --> 00:25:57
They're going to be some
finite numbers that are
331
00:25:57 --> 00:25:59
opposites of each other.
332
00:25:59 --> 00:26:05
Because we're no
longer isolated.
333
00:26:05 --> 00:26:21
So let's just write out delta S
to go backward is dq reversible
334
00:26:21 --> 00:26:32
over T, and that's minus dw
over T, not zero anymore.
335
00:26:32 --> 00:26:39
And where we're going is from
2V to V, we're compressing,
336
00:26:39 --> 00:26:41
and it's an ideal gas.
337
00:26:41 --> 00:26:59
So this is the same as p dV,
but it's p dV over T which
338
00:26:59 --> 00:27:02
is the same thing as
going from 2V to V.
339
00:27:02 --> 00:27:05
Now we're going to replace
this with RT over V, and
340
00:27:05 --> 00:27:07
the T's will cancel.
341
00:27:07 --> 00:27:15
So it's R dV over V
going from 2V to one V.
342
00:27:15 --> 00:27:23
So it's just R times
the log of 1/2.
343
00:27:23 --> 00:27:27
So now we've just calculated
the value of delta S by
344
00:27:27 --> 00:27:30
constructing this
reversible path.
345
00:27:30 --> 00:27:33
Now this is delta S backward.
346
00:27:33 --> 00:27:37
To do the compression, so of
course delta S forward is
347
00:27:37 --> 00:27:46
just the opposite of that.
348
00:27:46 --> 00:27:58
So, delta S forward is R log 2.
349
00:27:58 --> 00:28:12
Reassuringly, that's
a positive number.
350
00:28:12 --> 00:28:16
So this process that we
considered originally, this
351
00:28:16 --> 00:28:20
irreversibly process in this
isolated system, remember, for
352
00:28:20 --> 00:28:25
the irreversible expansion we
considered the system isolated,
353
00:28:25 --> 00:28:27
Happened spontaneously, and
of course you know
354
00:28:27 --> 00:28:30
that's the case.
355
00:28:30 --> 00:28:37
Now, I'll just mention, and
later on we'll go into this in
356
00:28:37 --> 00:28:40
considerably more detail, but
I'll just mention at this stage
357
00:28:40 --> 00:28:45
through a little bit of sort of
foreshadowing that everything
358
00:28:45 --> 00:28:48
we've discussed so far
about entropy is with
359
00:28:48 --> 00:28:50
a macroscopic picture.
360
00:28:50 --> 00:28:54
We started with heat engines
and Carnot cycles and
361
00:28:54 --> 00:28:56
macroscopic processes,
reversible and irreversible
362
00:28:56 --> 00:28:58
processes and so on.
363
00:28:58 --> 00:29:02
But there's a, and that's
certainly the way it was
364
00:29:02 --> 00:29:04
all formulated originally.
365
00:29:04 --> 00:29:06
Part of the power of
thermodynamics is that it
366
00:29:06 --> 00:29:12
doesn't depend on a specific
microscopic model of matter.
367
00:29:12 --> 00:29:15
That said though, we have
a pretty good microscopic
368
00:29:15 --> 00:29:16
model of matter.
369
00:29:16 --> 00:29:20
We know about atoms and
molecules, and in fact there is
370
00:29:20 --> 00:29:26
an entirely microscopic
formulation of entropy that has
371
00:29:26 --> 00:29:31
to do with disorder and the
number of available states,
372
00:29:31 --> 00:29:36
microscopic states
available to a system.
373
00:29:36 --> 00:29:41
So I'll just state what
we'll see in much
374
00:29:41 --> 00:29:44
more detail later on.
375
00:29:44 --> 00:29:57
Which is in microscopic terms,
it's going to turn out that the
376
00:29:57 --> 00:30:04
entropy of a system can be
given by R over Avogadro's
377
00:30:04 --> 00:30:16
number times the log of
the number of distinct
378
00:30:16 --> 00:30:24
microscopic states that are
available to a system.
379
00:30:24 --> 00:30:28
Right so in this isolated
system, when we double that
380
00:30:28 --> 00:30:33
volume, each molecule
suddenly has twice as many
381
00:30:33 --> 00:30:36
states available to it.
382
00:30:36 --> 00:30:38
You could formulate
that in a lot of ways.
383
00:30:38 --> 00:30:42
You can sort of divide up the
volume into tiny little
384
00:30:42 --> 00:30:46
molecule-sized cubes and
realize well, now, there
385
00:30:46 --> 00:30:49
are twice as many of them.
386
00:30:49 --> 00:30:52
Now if I have one molecule,
the number of states doubles.
387
00:30:52 --> 00:30:56
If I have n molecules, the
number of states goes
388
00:30:56 --> 00:30:57
up by 2 to the n.
389
00:30:57 --> 00:30:59
Because for each one of the
molecules, I can select any one
390
00:30:59 --> 00:31:04
of those states, and then the
next one I can select any one.
391
00:31:04 --> 00:31:05
So you have this enormous
increase in the
392
00:31:05 --> 00:31:08
number of states.
393
00:31:08 --> 00:31:19
So delta S in this case looks
like R over NA log of 2 to the
394
00:31:19 --> 00:31:22
power NA, it's Avogadro's
number, for a mole
395
00:31:22 --> 00:31:25
of molecules.
396
00:31:25 --> 00:31:30
This can come out
and cancel this.
397
00:31:30 --> 00:31:34
There's our R log 2 result.
398
00:31:34 --> 00:31:37
But the point I'm making here
is that could actually be
399
00:31:37 --> 00:31:42
derived from entirely
microscopic consideration.
400
00:31:42 --> 00:31:44
We haven't gone
through that yet.
401
00:31:44 --> 00:31:48
We've done entirely macroscopic
thermodynamics so far, but
402
00:31:48 --> 00:31:51
we'll do a little bit of
treatment of statistical
403
00:31:51 --> 00:31:56
mechanics, is what is called
the whole microscopic
404
00:31:56 --> 00:31:59
formulation of entropy
in thermodynamics.
405
00:31:59 --> 00:32:03
And that's a really important
fundamental result.
406
00:32:03 --> 00:32:08
And we'll get to it in much
more detail in a while.
407
00:32:08 --> 00:32:12
But for now, I just want to
continue calculating entropy
408
00:32:12 --> 00:32:31
changes for a few more
kinds of processes.
409
00:32:31 --> 00:32:37
So, how about if we have two
different substances and we
410
00:32:37 --> 00:32:50
mix them, entropy of mixing.
411
00:32:50 --> 00:33:00
So, we start with some number
of moles of substance A in some
412
00:33:00 --> 00:33:04
volume VA, and some other
number of moles of substance
413
00:33:04 --> 00:33:09
B in volume VB, separated.
414
00:33:09 --> 00:33:11
And then we remove the
barrier, of course we know
415
00:33:11 --> 00:33:14
they're going to mix.
416
00:33:14 --> 00:33:20
So we'll have some total number
of moles, nA plus nB, and
417
00:33:20 --> 00:33:23
they'll be filling some total
volume, which is the sum of
418
00:33:23 --> 00:33:26
the two original volumes.
419
00:33:26 --> 00:33:30
Certainly this is what we
expect to happen spontaneously.
420
00:33:30 --> 00:33:41
So are process is nA moles of
A, gas, initial volume VA and
421
00:33:41 --> 00:33:47
T, plus nB of substance B,
I think this is miswritten
422
00:33:47 --> 00:33:49
as A there in your notes.
423
00:33:49 --> 00:34:01
Gas VB and T goes to nA moles
of A plus nB moles of B.
424
00:34:01 --> 00:34:08
Gas, total volume and T.
425
00:34:08 --> 00:34:15
And we'll have constant
temperature and pressure.
426
00:34:15 --> 00:34:19
Well, there was no
p v worked done.
427
00:34:19 --> 00:34:22
The temperature didn't change
and it's ideal gases, so delta
428
00:34:22 --> 00:34:28
u is zero, but what
about delta S?
429
00:34:28 --> 00:34:31
Well again, to calculate it, we
need to find a reversible path.
430
00:34:31 --> 00:34:34
This is irreversible if we
just remove the barrier
431
00:34:34 --> 00:34:37
and let it all go.
432
00:34:37 --> 00:34:40
But we could at least imagine a
reversible path, and we could
433
00:34:40 --> 00:34:44
actually construct this
for the right substances.
434
00:34:44 --> 00:34:53
We could imagine that we can
find some sort of piston here.
435
00:34:53 --> 00:35:01
And we're going to prepare to
push it inward, with a membrane
436
00:35:01 --> 00:35:07
that's permeable only to b.
437
00:35:07 --> 00:35:14
And over here we could set up
the exact same thing, but this
438
00:35:14 --> 00:35:23
one's permeable only to a.
439
00:35:23 --> 00:35:27
And then we could just push
them to recover this state.
440
00:35:27 --> 00:35:29
And we can do it reversibly.
441
00:35:29 --> 00:35:33
So we could construct a
reversible process that
442
00:35:33 --> 00:35:36
does the reverse of this.
443
00:35:36 --> 00:35:40
And in this case, it'll be
isothermal still, constant
444
00:35:40 --> 00:35:45
pressure, reversible
compression of the two gases.
445
00:35:45 --> 00:36:22
So, reversible compression
and demixing.
446
00:36:22 --> 00:36:23
All right, so what happens?
447
00:36:23 --> 00:36:28
Well, of course, our delta S
of demixing is going to be
448
00:36:28 --> 00:36:34
minus delta S of mixing.
449
00:36:34 --> 00:36:40
Delta u of demixing
is still zero.
450
00:36:40 --> 00:36:47
Temperature didn't change,
ideal gases, and there's some
451
00:36:47 --> 00:36:52
reversible work and heat.
452
00:36:52 --> 00:37:04
So dw in this reversible case,
is just minus pA dVA minus pB
453
00:37:04 --> 00:37:12
dVB, the infinitesimal amount
of work done as these things
454
00:37:12 --> 00:37:15
are gradually moved
toward each other.
455
00:37:15 --> 00:37:27
So delta S for demixing, of
course it's dq reversible over
456
00:37:27 --> 00:37:30
T, but just like we did right
there, of course, in this case
457
00:37:30 --> 00:37:33
that's just negative dw.
458
00:37:33 --> 00:37:40
So it's integral from V to
VA, right, were compressing.
459
00:37:40 --> 00:37:43
So for substance a, we're
going to go back to there.
460
00:37:43 --> 00:37:45
It was occupying the whole
volume, and then it's going
461
00:37:45 --> 00:37:48
to end up in only part
of the volume, VA.
462
00:37:48 --> 00:37:54
pA dVA over T.
463
00:37:54 --> 00:38:09
Same thing for B, going to
volume B, pB dVB over T.
464
00:38:09 --> 00:38:14
And now we can substitute for
pressure, right? pV is nRT, so
465
00:38:14 --> 00:38:25
it's nA times R log of VA over
V plus nB of R log
466
00:38:25 --> 00:38:33
of VB over V.
467
00:38:33 --> 00:38:37
Now, so this is a suitable
answer, but I'm going to make
468
00:38:37 --> 00:38:42
if a little easier by putting
in terms of mole fractions.
469
00:38:42 --> 00:39:05
So, mole fractions, of course,
XA is nA over n, and XB is nB
470
00:39:05 --> 00:39:14
over n and XA is also equal to
VA over V and XB is equal to VB
471
00:39:14 --> 00:39:17
over V for ideal gases, right?
472
00:39:17 --> 00:39:18
The molar volumes are the same.
473
00:39:18 --> 00:39:21
We're at the same
temperature and pressure.
474
00:39:21 --> 00:39:32
So that immediately gives us
the delta X of demixing is nR
475
00:39:32 --> 00:39:42
times XA log XA plus XB log XB.
476
00:39:42 --> 00:39:48
And of course delta S of mixing
is the opposite of this, minus
477
00:39:48 --> 00:39:58
nR XA log XA plus XB log XB.
478
00:39:58 --> 00:40:00
Now XA and XB are
479
00:40:00 --> 00:40:01
mole fractions.
480
00:40:01 --> 00:40:06
They are between zero and
one, So their log rhythms
481
00:40:06 --> 00:40:08
are both negative.
482
00:40:08 --> 00:40:10
There's a negative sign.
483
00:40:10 --> 00:40:13
So delta S of mixing
is positive.
484
00:40:13 --> 00:40:15
That's reassuring.
485
00:40:15 --> 00:40:18
That tells us as we expect that
the mixing should happen
486
00:40:18 --> 00:40:28
spontaneously and irreversibly
when we remove the barrier.
487
00:40:28 --> 00:40:31
Any questions so far about
how we're going about these
488
00:40:31 --> 00:40:34
calculations of delta S?
489
00:40:34 --> 00:40:52
OK, let's just do a
couple more examples.
490
00:40:52 --> 00:40:54
Here's a really
straightforward one.
491
00:40:54 --> 00:40:58
What if we just heat stuff
up or cool it down.
492
00:40:58 --> 00:41:02
It happens all the time, but
it's pretty important, right?
493
00:41:02 --> 00:41:12
So let's just heat or
cool at constant volume.
494
00:41:12 --> 00:41:17
So, we're going to go from
substance A at T1 and some
495
00:41:17 --> 00:41:22
volume going to substance
A at T2 at some volume.
496
00:41:22 --> 00:41:25
We can do this.
497
00:41:25 --> 00:41:34
Delta S is integral of
dq reversible over T.
498
00:41:34 --> 00:41:43
Going from T1 to T2, but we
know how to do this, right?
499
00:41:43 --> 00:41:44
It's the heat that we need.
500
00:41:44 --> 00:41:47
It's just given by the constant
volume heat capacity times
501
00:41:47 --> 00:41:49
the change in temperature.
502
00:41:49 --> 00:42:00
So this is just T1 to
T2, Cv dT over T.
503
00:42:00 --> 00:42:11
And that is just Cv times the
log of T2 over T1 if Cv is
504
00:42:11 --> 00:42:16
temperature independent.
505
00:42:16 --> 00:42:18
That's not always the case.
506
00:42:18 --> 00:42:21
Usually if it's not too big an
excursion of temperature then
507
00:42:21 --> 00:42:25
it's a reasonable
approximation.
508
00:42:25 --> 00:42:33
So that's straightforward.
509
00:42:33 --> 00:42:42
By the way, notice that delta
S is greater than zero if
510
00:42:42 --> 00:42:44
T2 is greater than T1.
511
00:42:44 --> 00:42:45
1
512
00:42:45 --> 00:42:50
In other words, if we heated
it up, delta S is positive.
513
00:42:50 --> 00:42:54
Notice also delta S
is less than zero if
514
00:42:54 --> 00:42:57
T2 is less than T1.
515
00:42:57 --> 00:43:01
Delta S is negative
if we cooled it.
516
00:43:01 --> 00:43:05
Now, under certain conditions
we've seen that delta
517
00:43:05 --> 00:43:07
S can't be negative.
518
00:43:07 --> 00:43:09
Why can it be negative here?
519
00:43:09 --> 00:43:14
STUDENT: [UNINTELLIGIBLE].
520
00:43:14 --> 00:43:17
PROFESSOR NELSON:
Yes, I heard it.
521
00:43:17 --> 00:43:19
It's because the system
isn't isolated, right.
522
00:43:19 --> 00:43:23
It was for the isolated system
that delta S is always greater
523
00:43:23 --> 00:43:24
than or equal to zero.
524
00:43:24 --> 00:43:27
And you know, for the whole
universe entropy never
525
00:43:27 --> 00:43:30
decreases and so on.
526
00:43:30 --> 00:43:33
In this case, though, you know
we're taking a system and we're
527
00:43:33 --> 00:43:36
putting it in contact with a
cold bath and cooling it down.
528
00:43:36 --> 00:43:38
It's certainly not isolated.
529
00:43:38 --> 00:43:39
Heat is being exchanged.
530
00:43:39 --> 00:43:43
It's going from the system to
the surroundings, to the bath.
531
00:43:43 --> 00:43:44
So it's fine.
532
00:43:44 --> 00:43:48
Delta S can be negative.
533
00:43:48 --> 00:43:53
What would happen if we
calculated delta S of a new
534
00:43:53 --> 00:43:55
entire system consisting of
the original system
535
00:43:55 --> 00:43:58
plus the heat bath?
536
00:43:58 --> 00:44:00
What would delta S be then?
537
00:44:00 --> 00:44:07
We already kind of did that
with those two blocks, T1 and
538
00:44:07 --> 00:44:11
T2 in an isolated systems.
539
00:44:11 --> 00:44:13
What do you think is
going to happen?
540
00:44:13 --> 00:44:17
In other words, if we call our
system the stuff that is put in
541
00:44:17 --> 00:44:21
contact with the heat bath,
well OK, then we've seen delta
542
00:44:21 --> 00:44:24
S can be plus, can be
positive or negative.
543
00:44:24 --> 00:44:28
Now let's call our system the
original system plus the heat
544
00:44:28 --> 00:44:32
bath, and we'll put them all in
some isolating box, that's
545
00:44:32 --> 00:44:34
thermally isolating and so on.
546
00:44:34 --> 00:44:37
And now we'll put our original
system, which is now sort of a
547
00:44:37 --> 00:44:40
sub, part of the system, and
it's now going to be in
548
00:44:40 --> 00:44:45
contact with the cold bath.
549
00:44:45 --> 00:44:48
The bath that it's in contact
with is colder than it is.
550
00:44:48 --> 00:44:50
That's why its delta
S is less than zero.
551
00:44:50 --> 00:44:53
But what's delta S to be
for the entire new system
552
00:44:53 --> 00:44:58
including the cold bath?
553
00:44:58 --> 00:44:59
STUDENT: Wouldn't it
greater than zero?
554
00:44:59 --> 00:45:01
PROFESSOR NELSON: Yes, it's
going to be positive, because
555
00:45:01 --> 00:45:04
it's an isolated system
and something happened
556
00:45:04 --> 00:45:09
spontaneously in it, namely the
original system got cooler and
557
00:45:09 --> 00:45:12
presumably the heat bath got at
least a little bit warmer.
558
00:45:12 --> 00:45:15
And it happened spontaneously,
which immediately means delta S
559
00:45:15 --> 00:45:20
had to be positive for that
entire assembly, of you know
560
00:45:20 --> 00:45:25
the original system
plus the bath.
561
00:45:25 --> 00:45:32
Here's another important
sort of process.
562
00:45:32 --> 00:45:44
Reversible phase change, we'll
melt something or freeze it.
563
00:45:44 --> 00:45:50
How about water -- liquid
100 degrees Celsius
564
00:45:50 --> 00:45:54
one bar goes to water.
565
00:45:54 --> 00:46:01
Gas 100 degrees
Celsius one bar.
566
00:46:01 --> 00:46:05
Well in that case, you
know what the heat is.
567
00:46:05 --> 00:46:11
It's just the delta H of
vaporization, right,
568
00:46:11 --> 00:46:15
at constant pressure.
569
00:46:15 --> 00:46:17
So there's nothing really
much to calculate here.
570
00:46:17 --> 00:46:28
Delta S of vaporization for H2O
at 100 degrees Celsius is just
571
00:46:28 --> 00:46:36
q of vaporization over the
boiling temperature, which is
572
00:46:36 --> 00:46:43
delta H of vaporization over
the boiling temperature.
573
00:46:43 --> 00:46:45
So that was easy.
574
00:46:45 --> 00:46:49
One last thing.
575
00:46:49 --> 00:46:53
What if we want to calculate
the entropy of melting or
576
00:46:53 --> 00:46:56
some phase transition
not at equilibrium?
577
00:46:56 --> 00:46:59
In this case it's at
equilibrium because it's
578
00:46:59 --> 00:47:03
water and water vapor at
100 degrees Celsius and
579
00:47:03 --> 00:47:05
one atmosphere or one bar.
580
00:47:05 --> 00:47:08
So it's at its boiling point.
581
00:47:08 --> 00:47:14
But what if it isn't
at its boiling point?
582
00:47:14 --> 00:47:24
So what if instead, let's
take H2O, liquid, at minus
583
00:47:24 --> 00:47:29
ten degrees Celsius and
one bar so it's cold.
584
00:47:29 --> 00:47:31
It's going to freeze.
585
00:47:31 --> 00:47:36
So it's going to turn into
solid water at minus 10 degrees
586
00:47:36 --> 00:47:40
Celsius and one bar, we're
going to have it be isothermal.
587
00:47:40 --> 00:47:43
Now you know that's going
to happen spontaneously,
588
00:47:43 --> 00:47:52
and it's irreversible.
589
00:47:52 --> 00:47:56
Irreversible, which means we
can't just straight away
590
00:47:56 --> 00:48:00
calculate delta S along this
path because it's irreversible.
591
00:48:00 --> 00:48:04
But what we can do, just like
we've done before for making
592
00:48:04 --> 00:48:07
cycles, we can make some
other set of events that
593
00:48:07 --> 00:48:08
are all reversible.
594
00:48:08 --> 00:48:12
We can construct a reversible
pass that will get there.
595
00:48:12 --> 00:48:13
How do we do it?
596
00:48:13 --> 00:48:17
Well you know for the phase
change, for the freezing to be
597
00:48:17 --> 00:48:20
reversible, it has to happen at
zero degrees Celsius right.
598
00:48:20 --> 00:48:23
You know that's where
you have reversible,
599
00:48:23 --> 00:48:25
freezing and melting.
600
00:48:25 --> 00:48:28
Reversible equilibrium between
liquid and solid water.
601
00:48:28 --> 00:48:33
So surely that's got
to get involved here.
602
00:48:33 --> 00:48:39
H2O liquid zero degrees
Celsius, one bar, in
603
00:48:39 --> 00:48:44
equilibrium with H2O solid at
zero degrees Celsius, one bar.
604
00:48:44 --> 00:48:48
Whatever reversible path we
construct that's going to go
605
00:48:48 --> 00:48:52
from liquid to solid water,
somewhere along the set of
606
00:48:52 --> 00:48:55
steps, that better
be one of them.
607
00:48:55 --> 00:48:58
So now what we need to do is go
from liquid at minus 10 degrees
608
00:48:58 --> 00:49:02
Celsius and one bar to
liquid at zero degrees
609
00:49:02 --> 00:49:05
Celsius and one bar.
610
00:49:05 --> 00:49:08
That's called heating.
611
00:49:08 --> 00:49:09
We have to heat it.
612
00:49:09 --> 00:49:15
And here we have to cool it.
613
00:49:15 --> 00:49:18
And we've already seen all
these three processes, right.
614
00:49:18 --> 00:49:23
So for the phase change,
it's reversible now.
615
00:49:23 --> 00:49:26
This is just delta
heat of fusion.
616
00:49:26 --> 00:49:30
Great.
617
00:49:30 --> 00:49:33
Because this is now reversible.
618
00:49:33 --> 00:49:35
This is going to be reversible.
619
00:49:35 --> 00:49:37
This is going to be reversible.
620
00:49:37 --> 00:49:38
Great.
621
00:49:38 --> 00:49:43
So dq reversible, it's going to
be the heat capacity at
622
00:49:43 --> 00:49:46
constant pressure, the example
we did before it was constant
623
00:49:46 --> 00:49:51
volume for the liquid dT.
624
00:49:51 --> 00:49:59
Here, dq reversible is
going to be Cp solid, dT.
625
00:49:59 --> 00:50:01
Those won't be the same right?
626
00:50:01 --> 00:50:03
It's one thing to say the heat
capacity of something doesn't
627
00:50:03 --> 00:50:07
change over some small
excursion in temperature, but
628
00:50:07 --> 00:50:09
it's another thing when the
thing actually changes phase.
629
00:50:09 --> 00:50:11
The heat capacity of the
solid and the liquid
630
00:50:11 --> 00:50:14
won't be the same.
631
00:50:14 --> 00:50:18
So this we can just finish up
in a jiffy, because we're just
632
00:50:18 --> 00:50:23
going to add the three things.
633
00:50:23 --> 00:50:30
So delta S is delta S of
heating, minus, I think there's
634
00:50:30 --> 00:50:34
a plus that should be a minus
there, minus delta S of fusion
635
00:50:34 --> 00:50:36
because it's going in the
direction of freezing the
636
00:50:36 --> 00:50:38
way we've written it.
637
00:50:38 --> 00:50:42
Plus delta S of cooling.
638
00:50:42 --> 00:50:52
So integral from T1 to the
melting point of Cp of the
639
00:50:52 --> 00:50:59
liquid dT over T, minus
delta H of fusion over
640
00:50:59 --> 00:51:03
the melting temperature.
641
00:51:03 --> 00:51:07
Plus the integral going from
the heat of the temperature
642
00:51:07 --> 00:51:15
of melting to T1 of Cp
of the solid dT over T.
643
00:51:15 --> 00:51:18
That's it, all right?
644
00:51:18 --> 00:51:24
So delta S is minus delta
H of fusion over T.
645
00:51:24 --> 00:51:26
That's what it would be for
just the reversible phase
646
00:51:26 --> 00:51:30
change happening at
zero degrees Celsius.
647
00:51:30 --> 00:51:34
And then there's this
additional part which is,
648
00:51:34 --> 00:51:39
we can write it from T1
to melting point of Cp of
649
00:51:39 --> 00:51:47
the liquid, minus Cp of
the solid, dT over T.
650
00:51:47 --> 00:51:50
Now usually we can assume that
these will be temperature
651
00:51:50 --> 00:51:55
independent in
their own phases.
652
00:51:55 --> 00:52:00
So we can usually write this as
minus delta H of fusion over
653
00:52:00 --> 00:52:10
the melting temperature plus Cp
of the liquid minus Cp of the
654
00:52:10 --> 00:52:21
solid log of T of
fusion over T1.
655
00:52:21 --> 00:52:24
Any questions?
656
00:52:24 --> 00:52:25
What did we do?
657
00:52:25 --> 00:52:28
We constructed a reversible
path going from here to here
658
00:52:28 --> 00:52:31
and calculated delta S that
way, and of course that has
659
00:52:31 --> 00:52:36
to be the same as delta S in
the one irreversible step.
660
00:52:36 --> 00:52:40
All right, more on entropy and
its consequences for how to
661
00:52:40 --> 00:52:43
figure out what happened
spontaneously next time.
662
00:52:43 --> 00:52:44